Ta thấy : \(\left|x-2021\right|\ge0\forall x,\left|y-2022\right|\ge0\forall y\\ =>\left|x-2021\right|+\left|y-2022\right|\ge0\)

Mà theo đề : \(\left|x-2021\right|+\left|y-2022\right|\le0\)

=> \(\left\{{}\begin{matrix}x-2021=0\\y-2022=0\end{matrix}\right.=>\left(x;y\right)=\left(2021;2022\right)\)

tick mk nha các bn ơi

xạo vừa vừa thôi mấy mắm ơi, chtt đâu có đâu

\(\left(\sqrt{x-1}+\sqrt{3-x}\right)^2\le\left(1^2+1^2\right)\left(x-1+3-x\right)=4\\ \Leftrightarrow\sqrt{x-1}+\sqrt{3-x}\le2\\ y^2+2\sqrt{2020}y+2022=\left(y^2+2y\sqrt{2020}+2020\right)+2\\ =\left(y+\sqrt{2020}\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=3-x\\y+\sqrt{2020}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\sqrt{2020}\end{matrix}\right.\)

Vậy ...

ĐKXĐ: \(3\ge x\ge1\)

Áp dụng BĐT Bunhiacopski:

\(1\sqrt{x-1}+1\sqrt{3-x}\le\sqrt{\left(1^2+1^2\right)\left(x-1+3-x\right)}=\sqrt{2.2}=2\)

Mặt khác: \(y^2+2\sqrt{2020}y+2022=\left(y+\sqrt{2020}\right)^2+2\ge2\)

Nên để thõa mãn yêu cầu bài toán thì

\(\left\{{}\begin{matrix}\sqrt{x-1}=\sqrt{3-x}\\y+\sqrt{2020}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(tm\right)\\y=-\sqrt{2020}\end{matrix}\right.\)

\(P=\dfrac{1}{2021}\left(\dfrac{2021^2}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{2021}.\dfrac{\left(2021+1\right)^2}{x+y}=\dfrac{1}{2021}.\dfrac{2022^2}{\dfrac{2022}{2021}}=2022\)

\(P_{min}=2022\) khi \(\left(x;y\right)=\left(1;\dfrac{1}{2021}\right)\)

sao cái đoạn \(\dfrac{1}{2021}\left(\dfrac{2021^2}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{2021}.\dfrac{\left(2021+1\right)^2}{x+y}\) làm kiểu gì ra thầy :)

O biết

Sửa đề: \(\frac{6}{\left(x-2\right)^2+2}=\left|y-2022\right|+\left|y-2025\right|\)

Ta có: \(\left(x-2\right)^2+2\ge2\forall x\)

=>\(\frac{6}{\left(x-2\right)^2+2}\le\frac62=3\forall x\)

\(\left|y-2022\right|+\left|y-2025\right|=\left|y-2022\right|+\left|2025-y\right|\ge\left|y-2022+2025-y\right|=3\forall y\)

mà \(\frac{6}{\left(x-2\right)_{}^2+2}=\left|y-2022\right|+\left|y-2025\right|\)

nên \(\frac{6}{\left(x-2\right)^2+2}=\left|y-2022\right|+\left|y-2025\right|=3\)

=>\(\begin{cases}\left(x-2\right)^2+2=\frac63=2\\ \left(y-2022\right)\left(y-2025\right)\le0\end{cases}\Rightarrow\begin{cases}x-2=0\\ 2022\le y\le2025\end{cases}\)

=>\(\begin{cases}x=2\\ y\in\left\lbrace2022;2023;2024;2025\right\rbrace\end{cases}\)

\(\Rightarrow2019\left|x-1\right|+2020\left|y-2\right|+2021\left|y-3\right|+2022\left|y-4\right|=2020+2022\)

\(\Rightarrow\hept{\begin{cases}\left|y-2\right|=1\\\left|x-1\right|=0\\\left|y-4\right|=1\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)